Answer to Question #64548 in General Chemistry for Jack

Question

Using heat formations , calculate the heat of vaporization for methyl alcohol, CH3OH, at 25 Celsius (answer:38 kj/mol)

How many kJ's are required to vaporize 28.6 g CH3OH at standard conditions (answer:33.9kj/mol)

i dont get how to do the work

Answer 1

Enthalpy of formation for liquid CH3OH is -238.57 kJ/mol and that for a gaseous CH3OH is -201.00 kJ/mol. Thus, the difference is -238.57 - (-201.00) = - 37.57 kJ/mol.
Heat of formation has the opposite value +37.57 kJ/mol.

28.6 g of CH3OH is 0.89 mol. Thus, to evaporate 28.6 g we need 37.57 kJ/mol * 0.89 mol = 33 kJ of heat.